Hi actually when i execute the code i get an output it will be spitted in echo statement but if i put echo in textarea i get only last value of string from my database
$sql = mysqli_query($con, "SELECT naveen from kumar where id<5");
while ($row = mysqli_fetch_array($sql)) {
$myString = $row['admin'];
$myArray = explode(',', $myString);
foreach ($myArray as $my_Array) {
echo $my_Array.'<br>';
}
}
?>
<form method="post" action="">
<textarea name="valid" cols="60" rows="5"><?php echo $my_Array;?></textarea>
</form>
<?php ?>
This is my result
You need to concatenate your result with previous one. Delcare $my_Array = ""; before while loop For break line in textarea you have to use
foreach ($myArray as $my_Array) {
$my_Array .= '
';
echo $my_Array; //IF you also want to echo here
}
You should paint a <textarea> inside of first bucle (while).
echo '<form method="post" action="">';
$sql = mysqli_query($con, "SELECT naveen from kumar where id<5");
while ($row = mysqli_fetch_array($sql)) {
$myString = $row['admin'];
$myArray = explode(',', $myString);
echo '<textarea name="valid" cols="60" rows="5">';
foreach ($myArray as $my_Array) {
echo = $my_Array.'<br>';
}
echo '</textarea>';
}
?>
</form>
I hope this example will solve your issue. Please let me know if you still have any queries.
<?php
$myvalue = array('one','two','three');
$myterm = implode('-', $myvalue);
$mynewvalue = explode('-',$myterm);
?>
<form method="post" action="">
<textarea name="valid" cols="60" rows="5">
<?php
foreach($mynewvalue as $mvn)
{
echo $mvn.',';
}
?>
</textarea>
</form>
You have to do it like this:
<form method="post" action="">
<?php $sql = mysqli_query($con, "SELECT naveen from kumar where id<5");
while ($row = mysqli_fetch_array($sql)) {
$myString = $row['admin'];
$myArray[] = explode(',', $myString);
}
?>
<textarea name="valid" cols="60" rows="5"><?php echo implode(" ",$my_Array);?></textarea>
</form>
<?php ?>
Related
I'm having a hard time displaying the title and note on each row of my database. I want to display one row (with the title and note) after each
from a form in a page heading to the displaying of row datas page.
This is my code below:
//
Let's say that we have 4 rows of datas. In my code, I can only display the first row, because it keeps having the first row's data. This is because the form is in the first php file. Then after I submit the form, it's directed to this file, and it keeps getting the first row.
<?php $con=mysqli_connect("localhost","root","","task");?>
<?php $results = mysqli_query($con, "SELECT * FROM note"); ?>
<?php while ($row = mysqli_fetch_array($results)) { ?>
<?php
$id=$row['id'];
echo ' ';
echo '<button class="call_modal" data-id="$id" style="cursor:pointer;">'. $row['title'] . '</button>';
?>
<?php
}?>
<?php $results = mysqli_query($con, "SELECT * FROM note"); ?>
<?php while ($row = mysqli_fetch_array($results)) { ?>
<div class="note" data-id="<?= $row['id'] ?>">
<div class="modal">
<div class="modal_close close"></div>
<div class="modal_main">
<?php
echo '<br><br>';
echo '<div class="padding">'.$row['title'].'';
echo '<br><br><br><br>';
echo ''.$row['note'].'</div>';
?>
</div>
</div>
<?php
}?>
<?php
<?php
function test_input($data) {
$data = trim($data);
$data = stripslashes($data);
$data = htmlspecialchars($data);
return $data;
} ?>
The code below is not tested but it should be correct and give you a better idea of what you are doing right/wrong. I hope it helps..
<?php
$con=mysqli_connect("localhost","root","","task");
$results = mysqli_query($con, "SELECT * FROM note");
while ($row = mysqli_fetch_array($results)) { //starting your data row loop
$id=$row['id'];// you are creating a variable here but not using it two lines down.
echo ' ';
//YOUR OLD CODE echo '<button class="call_modal" data-id="$id" style="cursor:pointer;">'. $row['title'] . '</button>';
echo '<button class="call_modal" data-id="$id" style="cursor:pointer;">'. $id . '</button>';// use the variable "$id" that you created here
// You dont need the next two lines, you already did a query and have the data loaded in the "$results" array
/* $results = mysqli_query($con, "SELECT * FROM note");
while ($row = mysqli_fetch_array($results)) */
?> // close the php tag if you are going to switch to html instead of "echoing html"
/* OLD CODE <div class="note" data-id="<?= $row['id'] ?>"> you were missing the "php" in the php tags */
<div class="note" data-id="<?php echo $id; ?>"> // corrected code
<div class="modal">
<div class="modal_close close"></div>
<div class="modal_main">
<?php //switching back to php so create your open tag again...
echo '<br><br>';
echo '<div class="padding">'.$row['title'].'';
echo '<br><br><br><br>';
echo ''.$row['note'].'</div>';// you dont NEED the '' before .$row.... unless you want it but its just essentially a blank string
?>
</div>
</div>
<?php
} // ending your "for each data row" here
?>
<?php
// PS you're not using this function anywhere unless its in ommited code?
function test_input($data) {
$data = trim($data);
$data = stripslashes($data);
$data = htmlspecialchars($data);
return $data;
} ?>
In your loop you're using mysqli_fetch_array, which returns an array with each element in that array containing the field value.
What you want is mysqli_fetch_assoc instead, this will return a hash which you can then use the way you're using it now.
Another thing is that you don't need to have 2 loops in there querying the database. And please indent your code, it makes it really hard for you and everyone else to read it.
Here is the updated/cleaned up version of your code. This has been tested, you can find a sample code with instructions to run on my Github here.
<?php
$con = mysqli_connect("localhost", "root", "", "task");
$results = mysqli_query($con, "SELECT * FROM note");
while ($row = mysqli_fetch_assoc($results)) {
$id = $row['id'];
echo ' ';
echo '<button class="call_modal" data-id="' . $id . '" style="cursor:pointer;">'. $row['title'] . '</button>';
?>
<div class="note" data-id="<?= $row['id'] ?>">
<div class="modal">
<div class="modal_close close"></div>
<div class="modal_main">
<?php
echo '<br /><br />';
echo '<div class="padding">' . $row['title'];
echo '<br /><br /><br /><br />';
echo $row['note'];
echo '</div>'
?>
</div>
</div>
</div>
</div>
<?php
}
?>
<?php
function test_input($data) {
$data = trim($data);
$data = stripslashes($data);
$data = htmlspecialchars($data);
return $data;
}
I am trying to pass hidden value from page to another page and it work fine only for first record however for other records it's showing error
Here is the code:
$sql = "SELECT id,jdes,title FROM job";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
?>
<input type="hidden" id="hidden_user_id" value="<?php echo $row["id"] ?>">
<h3><?php echo $row["title"] ?>:</h3>
<p class="lead">
<?php echo $row["jdes"] ?>
</p>
<button type="button" id="requestthis" class="btn btn-primary">
Request
</button>
<?php
}
} else {
echo "Nothing to display Yet";
}
?>
jobs-inner.php
<?php
echo $_GET['hidden_id'];
?>
Javascript:-
$(function() { //ready function
$('#requestthis').on('click', function(e){ //click event
e.preventDefault();
var hidden_id = $('#hidden_user_id').val();
var url = "jobs-inner.php?hidden_id="+hidden_id;
window.location.replace(url);
})
})
Error:-
Undefined index: hidden_id in C:\wamp64\www\project\jobs-inner.php on line 3
It might be a simple problem but I am a beginner and I can't figure it out.
Your value is unique but the id isn't. Make the id of the input unique something like below.
<input type="hidden" id="hidden_user_<?php echo $row["id"] ?>" value="<?php echo $row["id"] ?>">
but you would have to do a count on code below to make it display base on how many rows you have.
<?php
echo $_GET['hidden_id'];
?>
Without JavaScript
$sql = "SELECT id,jdes,title FROM job";
$result = $conn->query($sql);
$count = 1;
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
?>
<input type="hidden" id="hidden_user_<?php echo $count ?>" value="<?php echo $row["id"] ?>">
<h3><?php echo $row["title"] ?>:</h3>
<p class="lead"><?php echo $row["jdes"] ?></p>
<form id="<?php echo $count ?>" action="jobs-inner.php?hidden_id=<?php echo $row["id"] ?>" method="post">
<input type="submit" vaule="Request">
</form>
<?php
$count++;
}
} else {
echo "Nothing to display Yet";
}
?>
My code:
<?php
require_once 'core/init.php';
if($result = $db->query ("SELECT * from countries")) {
if($count = $result->num_rows) {
$rows = $result->fetch_all(MYSQLI_ASSOC);
}
}
?>
<div class="region">
<h1>Europe</h1>
<ul class="country" style ="list-style-type:none">
<?php foreach ($rows as $key => $row) { ?>
<li>
<a href=""=<?php echo $row['country_id']; ?>">
<a href=""=<?php echo $row['region_id']; ?>">
<?php echo $row['country_name']; ?></a> <br>
<?php
if (isset($_FILES['country_img']) === true) {
if (empty($_FILES['country_img']['name']) === true) {
echo 'please choose a file!';
print_r ($_POST);
} else {
$allowed = array('jpg', 'jpeg', 'gif', 'png');
$file_name = $_FILES['country_img']['name'];
$file_extn = end(explode('.', $file_name));
$file_temp = $_FILES['country_img']['tmp_name'];
if (in_array($file_extn, $allowed) ===true) {
if ($_POST['id'] = $_POST['id']) {
addcountryimage($row['country_id'], $file_temp, $file_extn);
}
header ('Location: tes2.php');
break;
}else {
echo 'Incorrect file type. Allowed: ';
echo implode (', ', $allowed);
}
}
}
if (empty($row['country_img'] === false)) {
echo '<img src="', $row['country_img'], '" alt="', $row['country_name'], '\'s Image">';
}
?>
<p> <?php echo $row['country_bio']; ?></p>
<?php
global $session_user_id;
if (is_admin($session_user_id) === true) { ?>
<form action="" method="POST" name="" enctype="multipart/form-data">
<input type="hidden" name="id" id="hidden_id">
<input type="file" name="country_img"> <input type="submit" onclick="document.getElementById('hidden_id').value=$row['country_id']" />
</form>
<?php } ?>
<p> Tour to <?php echo $row['country_name']; ?></p>
</li>
<?php } ?>
</ul>
<h3> More European Country </h3>
</div>
Everything above works, except the fact that it can't differentiate between each submit button. For example, when I update the image on the 5th forms it will always update the 1st form. How can I do this?
The origin of your issue is using multiple id attributes with the same value in your JavaScript events, this is regarded wrong, every element id must be unique, beside this, onclick event with submit typed inputs of form is not working properly.
Add an incremental counter to your loop and then use to distinguish each form hidden element id.
<?php $i = 0;?>
<?php foreach ($rows as $key => $row) { ?>
....
<input type="hidden" name="id" id="hidden_id_<?php echo $i;?>">
....
<input type="submit" onclick="document.getElementById('hidden_id_<?php echo $i;?>').value=$row['country_id']" />
<?php
$i++;
} ?>
Another note, I think that onclick event does not works fine with submit type inputs, so I suggest to replace it with onsubmit event in the form itself as:
<form action="" method="POST" name="" enctype="multipart/form-data" onsubmit="document.getElementById('hidden_id_<?php echo $i;?>').value=$row['country_id']">
I need the code for getting the addr and pnum of the patient when I choose the pname in the combo box.
How can I do that?
<script>
function getVal() {
document.getElementById("text2").value = document.getElementById("model").value;
}
</script>
<body>
//code in opening and getting my addr and pnum in dbase
<?php
include('connect.php');
$pname=$_GET['tpname'];
$result = mysql_query("SELECT * FROM tblnpatient where pname='$pname'");
while($row = mysql_fetch_array($result))
{
$pnum=$row['pnum'];
$addr=$row['addr'];
}
?>
//code for choosing pname
<tr><td>Patient Name:
<div id="ma">
<select name="pname" class="textfields" id="model" style="width:180px;" onchange="getVal();">
<option id="0" >--Select Patient Name--</option>
<?php
$con=mysqli_connect("localhost","root","","dbnpatient");
if (mysqli_connect_errno()){
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$pnum=$_GET['pnum'];
$query = mysqli_query($con, "SELECT * FROM tblnpatient");
while($row = mysqli_fetch_array($query)){
$pnum = $row['pnum'];
$pname = $row['pname'];
?>
<option id="<?php echo $pnum; ?>" value="<?php echo $pname; ?>"><?php echo $pname; ?> </option>
<?php } ?>
</select>
//code for getting pname and addr
Address:<input type="text" name="ename" size="20" id="ma" value="<?php echo $addr ?>"/>
Name:<input type="text" name="ename" size="20" id="ma" value="<?php echo $pname ?>"/>
In your while loop add the following (if you have that column otherwise replace with something similar)
$paddress = $row['paddress'];
You can store the needed information by using the data attribute in your options
<option id="<?php echo $pnum; ?>" data-pname="<?php echo $pname; ?>" data-paddress="<?php echo $paddress; ?>" value="<?php echo $pname; ?>"><?php echo $pname; ?></option>
Then change your getVal() function
function getVal() {
var options = document.getElementById("model").options;
if (options.selectedIndex != -1) {
var addr = document.getElementById('paddress');
var name = document.getElementById('pname');
addr.value = options[options.selectedIndex].getAttribute('data-paddress');
name.value = options[options.selectedIndex].getAttribute('data-pname');
}
}
Now change the id's of the input fields for the address and the name to paddress and pname. It is important to know to never have duplicate id's
I hope that helped
Good morning! Please some body help or give some sample code about how to read text value (per line)
<body onLoad="displayResult()">
<table align="center">
<tr>
<td>
<?php $query2="SELECT * FROM upload1 WHERE NAME='xmlsample1.xml'";
$result1=mysql_query($query2);
$row = mysql_fetch_array($result1);
?>
<form action="">
<textarea id="validxml" rows="50" cols="100">
<?php echo $row['CONTENT']; ?>
</textarea>
<br><br>
<input type="button" value="Verify XML" onClick="validateXML()" />
</form>
</td>
</tr>
</table>
</body>
Access the value of textareaand split newline
console.log(document.getElementById('validxml').value.split("\n"));
JSFiddle
connect.php should be kept on a separate, secure web page.
<?php
function db(){
return new mysqli('host', 'username', 'password', 'database');
}
?>
Now on your other page:
<?php
include 'connect.php'; $db = db();
$sq = $db->query("SELECT * FROM upload1 WHERE name='xmlsample1.xml'"); $ta = '';
if($sq->num_rows > 0){
while($row = $sq->fetch_object()){
$ta .= "<textarea class='validxml' name='{$row->name}'>{$row->content}</textarea>";
}
}
$sq->free(); $db->close();
?>
Now just echo $ta into your HTML. Of course, I wouldn't even use a <textarea> to just output code, unless you want your user to be able to edit it.